Problem: On the way to school, Karen became fixated on the puzzle game on her phone!

<image>

The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number 0.

One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.

To win the level, after all the moves, the number in the cell at the i-th row and j-th column should be equal to gi, j.

Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task!

Input

The first line of input contains two integers, n and m (1 ≤ n, m ≤ 100), the number of rows and the number of columns in the grid, respectively.

The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains gi, j (0 ≤ gi, j ≤ 500).

Output

If there is an error and it is actually not possible to beat the level, output a single integer -1.

Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.

The next k lines should each contain one of the following, describing the moves in the order they must be done:

  * row x, (1 ≤ x ≤ n) describing a move of the form "choose the x-th row". 
  * col x, (1 ≤ x ≤ m) describing a move of the form "choose the x-th column". 



If there are multiple optimal solutions, output any one of them.

Examples

Input

3 5
2 2 2 3 2
0 0 0 1 0
1 1 1 2 1


Output

4
row 1
row 1
col 4
row 3


Input

3 3
0 0 0
0 1 0
0 0 0


Output

-1


Input

3 3
1 1 1
1 1 1
1 1 1


Output

3
row 1
row 2
row 3

Note

In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:

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In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.

In the third test case, Karen has a grid with 3 rows and 3 columns. She can perform the following 3 moves to beat the level:

<image>

Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3.

Solution (in PYTHON3):

Solution:
n, m = map(int,input().split())
table = list()
table_T = list()
for i in range(n):
    table.append(list(map(int,input().split())))

for i in range(m):
    a = [table[j][i] for j in range(n)]
    table_T.append(a)

OUT = list()

if n <= m:
    a = 0
    while a < n:
        is_0 = False
        for i in table[a]:
            if i == 0:
                is_0 = True
                break
        if is_0 == False:
            b = min(table[a])
            OUT.extend(['row {0}'.format(a+1)]*b)
            for i in range(m):
                table[a][i] -= b
                table_T[i][a] -= b
        else:
            a += 1

    a = 0
    while a < m:
        is_0 = False
        for i in table_T[a]:
            if i == 0:
                is_0 = True
                break
        if is_0 == False:
            b = min(table_T[a])
            OUT.extend(['col {0}'.format(a+1)]*b)
            for i in range(n):
                table[i][a] -= b
                table_T[a][i] -= b
        else:
            a += 1
else:
    a = 0
    while a < m:
        is_0 = False
        for i in table_T[a]:
            if i == 0:
                is_0 = True
                break
        if is_0 == False:
            b = min(table_T[a])
            OUT.extend(['col {0}'.format(a + 1)]*b)
            for i in range(n):
                table[i][a] -= b
                table_T[a][i] -= b
        else:
            a += 1

    a = 0
    while a < n:
        is_0 = False
        for i in table[a]:
            if i == 0:
                is_0 = True
                break
        if is_0 == False:
            b = min(table[a])
            OUT.extend(['row {0}'.format(a + 1)] * b)
            for i in range(m):
                table[a][i] -= b
                table_T[i][a] -= b
        else:
            a += 1

fail = False
for i in range(n):
    if sum(table[i]) != 0:
        fail = True
        break

if fail != True:
    print(len(OUT))
    for i in OUT:
        print(i)
else:
    print(-1)